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Chapter 4: Problem 37

Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \frac{x}{x+1} $$

Short Answer

Expert verified
The derivative is \( f'(x) = \frac{1}{x(x+1)} \).

Step by step solution

01

Express the Function in Terms of Natural Logarithms

The function given is \( f(x) = \ln \frac{x}{x+1} \). Notice that \( \ln \) denotes the natural logarithm, which is a hint that no conversion is necessary. However, we can use the property of logarithms to simplify this expression. The property \( \ln \frac{a}{b} = \ln a - \ln b \) can be applied: \( f(x) = \ln x - \ln (x+1) \).
02

Differentiate Each Term Separately

We need to differentiate \( f(x) = \ln x - \ln (x+1) \) with respect to \( x \). Use the differentiation formula for natural logarithms: \( \frac{d}{dx} \ln u = \frac{1}{u} \frac{du}{dx} \). For \( \ln x \), \( \frac{d}{dx} \ln x = \frac{1}{x} \). For \( \ln (x+1) \), using \( u = x+1 \), we have \( \frac{d}{dx} \ln (x+1) = \frac{1}{x+1} \).
03

Combine the Derivatives

Combine the derivatives from Step 2 to find the derivative of \( f(x) \): \( f'(x) = \frac{d}{dx} \left( \ln x - \ln (x+1) \right) = \frac{1}{x} - \frac{1}{x+1} \).
04

Simplify the Expression

Simplify the expression for the derivative: \( f'(x) = \frac{x+1 - x}{x(x+1)} = \frac{1}{x(x+1)} \). This is a simplified form of the derivative of the given function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithm
The natural logarithm is denoted as \( \ln \) and it's a fundamental concept in calculus. It refers to a logarithm with base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. The natural logarithm has distinct properties which make it particularly useful in calculus and differential equations.
  • The equation \( \ln(e) = 1 \) signifies that the natural logarithm of \( e \) itself is always 1.
  • Natural logarithms allow for straightforward differentiation and integration, contributing to their frequent use in calculus.
When working with natural logarithms, remember that they can transform multiplication into addition and division into subtraction, making seemingly complex expressions easier to work with. This property is leveraged in simplifying and differentiating functions, as shown in the original problem where \( \ln \frac{x}{x+1} = \ln x - \ln(x+1) \).
Derivative
A derivative represents the rate at which a function is changing at any given point. It's a cornerstone of calculus. The derivative of a function at a certain point provides the slope of the tangent line to the graph of the function at that same point. The process of finding a derivative is known as differentiation.
  • The basic rule for differentiation of natural logarithms is \( \frac{d}{dx} \ln u = \frac{1}{u} \cdot \frac{du}{dx} \), with \( u \) being a function of \( x \).
  • The simplest example is \( \frac{d}{dx} \ln x = \frac{1}{x} \), showing that the derivative of \( \ln x \) is a simple reciprocal function.
In the exercise, differentiating \( \ln x \) and \( \ln(x+1) \) separately and then combining them gives you the derivative of the entire logarithmic expression. Mastering derivatives assists in understanding functions' behaviors, including their increases, decreases, and points of inflection.
Properties of Logarithms
Understanding the properties of logarithms can significantly simplify complex expressions. These properties are crucial in transforming multiplication into addition or division into subtraction, providing a more straightforward form to differentiate.
  • The change of base formula: \( \log_{b}a = \frac{\ln a}{\ln b} \), although not needed in this problem, is essential to know.
  • Another fundamental property used was \( \ln \frac{a}{b} = \ln a - \ln b \), which makes expressions more manageable by reducing fractions to differences.
As in the given exercise, using the property \( \ln \frac{x}{x+1} = \ln x - \ln(x+1) \) allowed us to break down the expression into manageable pieces for differentiation. Knowing these properties helps address many calculus problems by reducing the complexity of logarithmic expressions.
Chain Rule
The chain rule is a powerful method in calculus used to differentiate composite functions. If a function is composed of two functions, say \( g(x) \) and \( h(x) \), the chain rule provides a way to find the derivative of this composition \( f(x) = g(h(x)) \). The rule states:\[\frac{d}{dx} g(h(x)) = g'(h(x)) \cdot h'(x)\]
  • This rule is vital when you have to differentiate a function containing another function, such as \( \ln( x + 1 ) \), where \( h(x) = x+1 \).
  • By applying the chain rule, you can tackle problems where functions are nested within each other.
In the provided example, while differentiating \( \ln(x + 1) \), it's essential to recognize \( x + 1 \) as \( u \), implying it must be differentiated as an internal function using the chain rule. Understanding the chain rule allows you to differentiate such compound structures efficiently, expanding your calculus toolkit significantly.

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Most popular questions from this chapter

Find the derivatives of the following functions: $$ f(x)=\cos ^{2}\left(2 x^{2}+3\right) $$

(a) Use the formal definition to find the derivative of \(y=\) \(1-x^{3}\) at \(x=2\) (b) Show that the point \((2,-7)\) is on the graph of \(y=1-x^{3}\), and find the equation of the normal line at the point \((2,-7)\). (c) Graph \(y=1-x^{3}\) and the tangent line at the point \((2,-7)\) in the same coordinate system.

(a) Use the formal definition of the derivative to find the derivative of \(y=2 x^{2}\) at \(x=-1\). (b) Show that the point \((-1,2)\) is on the graph of \(y=2 x^{2}\), and find the equation of the tangent line at the point \((-1,2)\). (c) Graph \(y=2 x^{2}\) and the tangent line at the point \((-1,2)\) in the same coordinate system.

Find \(c\) so that \(f^{\prime}(c)=0 .\) \(f(x)=x^{2}+6 x+9\)

Calculate the linear approximation for \(f(x)\) : $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=e^{x-1}\) at \(a=1\)
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